Question

Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(– 2, – 16), and x intercepts at x = -6 and x = 2. (Do not include the negative sign in your answer.)

y = x^2 + 4x − ___

Answer 1

You would use the technique of completing the square for the vertex form of a parabola:

y = a(x - h)^2 + k where (h, k) is the vertex

Since the partial solution has an understood "1" as the coefficient on x^2, we have the simpler equation:

y = (x - h)^2 + k -> y = [x - (-2)]^2 + (-16)

y = (x + 2)^2 - 16 -> y = x^2 + 4x + 4 - 16

y = x^2 + 4x - 12

Answer 2

All good now, @Melxo52 ?

Answer 3

why thank you, you should be my math teacher! lol